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olga2289 [7]
2 years ago
12

If 6.49 mol of ethane (C2H6) undergo combustion according to the unbalanced equation

Chemistry
1 answer:
MArishka [77]2 years ago
3 0

Answer:

22.715 moles of oxygen are used

Explanation:

Given data:

Number of moles of ethane = 6.49 mol

Number of moles of O₂ required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂       →     4CO₂ + 6H₂O

Now we will compare the moles of oxygen with ethane.

                    C₂H₆           :           O₂  

                       2              :             7

                     6.49           :           7/2×6.49 = 22.715 mol

Thus, 22.715 moles of oxygen are used.

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3 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
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Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

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3 years ago
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